On Orthocentric Systems in Strictly Convex Normed Planes

E extracta mathematicae Vol. 24, N´um.1, 31 – 45 (2009)

Horst Martini, Senlin Wu ∗

Faculty of Mathematics, Chemnitz University of Technology, 09107 Chemnitz, Germany [email protected] Department of Applied Mathematics, Harbin University of Science and Technology 150080 Harbin, China, senlin [email protected]

Presented by Pier L. Papini Received November 12, 2008

Abstract: It has been shown that the three-circles theorem, which is also known as T¸it¸eica’s or Johnson’s theorem, can be extended to strictly convex normed planes, with various applications. From this it follows that the notions of orthocenters and orthocentric systems in the Euclidean plane have natural analogues in strictly convex normed planes. In the present paper (which can be regarded as continuation of [5] and [14]) we derive several new characterizations of the Euclidean plane by studying geometric properties of orthocentric systems in strictly convex normed planes. All these results yield also geometric characterizations of inner product spaces among all real Banach spaces of dimension ≥ 2 having strictly convex unit balls. Key words: Birkhoﬀ orthogonality, Busemann angular bisector, C-orthocenter, inner product space, isosceles orthogonality, Minkowski plane, normed linear space, orthocenter, orthocentric system, three-circles theorem. AMS Subject Class. (2000): 46B20, 46C15, 52A10.

1. Introduction

By (M2, C) we denote an arbitrary normed or Minkowski plane (i.e., a real two-dimensional normed linear space) with unit circle C, origin O, and norm k·k. We restrict our discussion to strictly convex Minkowski planes, whose unit circles are strictly convex curves with midpoint O, thus not containing a non-degenerate segment. Basic references to the geometry of Minkowski planes and spaces are [18], [16], [17], and the monograph [19]. For any point x ∈ (M2, C) and any number λ > 0, the set C(x, λ) := x+λC is said to be the circle centered at x and having radius λ. For x 6= y, we denote by hx, yi the line passing through x and y, by [x, y] the segment between x and y, and by [x, yi the ray with starting point x passing through y. The

∗This work is partially supported by NSFC (China), grant 10671048.

31 32 h. martini, s. wu distance from a point x to a non-empty set A is denoted by d(x, A), i.e., d(x, A) = inf{kz − xk : z ∈ A}. It has been shown by E. Asplund and B. Gr¨unbaum in [5] that the following theorem, which is the extension of the classical three-circles theorem in the Euclidean plane, also holds in strictly convex, smooth Minkowski planes. (The three-circles theorem is also called T¸it¸eica’s or Johnson’s theorem; see the survey [13] and the monograph [12, p. 75]. For the related concept of orthocentricity, even in Euclidean n-space with n ≥ 2, we refer to the survey contained in [9] and to [7].)

Theorem 1.1. If three circles C(x1, λ), C(x2, λ), and C(x3, λ) pass through a common point p4 and intersect pairwise in the points p1, p2, and p3, then there exists a circle C(x4, λ) such that {p1, p2, p3} ⊆ C(x4, λ).

Actually, the smoothness condition can be relaxed, i.e., Theorem 1.1 holds in strictly convex Minkowski planes (cf. [19, Theorem 4.14, p. 104] and [14]). This theorem is also basic for extensions of Cliﬀord’s chain of theorems to strictly convex normed planes; see [15].

The point p4 in Theorem 1.1 is called the C-orthocenter of the triangle p1p2p3, and by Theorem 1.1 it is also evident that pi is the C-orthocenter of the triangle pjpkpl, where {i, j, k, l} = {1, 2, 3, 4}. Thus it makes sense to call a set of four points, each of which is the C-orthocenter of the triangle formed by the other three points, a C-orthocentric system. See [20] for another concept of orthocenters which is related to Birkhoﬀ orthogonality. The proof of the above theorem for strictly convex Minkowski planes is based on the following properties of strictly convex Minkowski planes (cf. [5] and [18, p. 106]), which will be used throughout the paper:

Lemma 1.2. Let (M2, C) be a strictly convex Minkowski plane. If x1 6= x2 and {y1, y2} ⊆ C(x1, λ) ∩ C(x2, λ), then x1 + x2 = y1 + y2.

Lemma 1.3. Any three non-collinear points in a strictly convex Minkowski plane are contained in at most one circle.

The following facts are well known in Euclidean geometry:

1. The three altitudes of a triangle intersect in a point called orthocenter of that triangle. on orthocentric systems 33

2. The altitude to the base of an isosceles triangle bisects the corresponding vertex angle.

3. If one of the altitudes of a triangle is also an angular bisector, then the triangle is isosceles.

4. The altitude to the base of an isosceles triangle bisects its base.

5. If one of the altitudes of a triangle is also a median (i.e., a segment from a vertex to the midpoint of the opposite side), then the triangle is isosceles.

As the notion of C-orthocenter can be viewed as a natural extension of that of orthocenter in Euclidean geometry, one may ask whether results related to orthocenters in Euclidean geometry still hold in Minkowski geometry in the sense of C-orthocenters. It is our aim to continue the investigations from [14] in this spirit. For our discussion we need the definitions of isosceles orthogonality, Birkhoff orthogonality, and Busemann angular bisectors. Let x, y ∈ (M2, C). The point x is said to be isosceles orthogonal to y if kx + yk = kx − yk, and in this case we write x ⊥I y (cf. [10]). On the other hand, x is said to be Birkhoff orthogonal to y if kx + tyk ≥ kxk holds for all t ∈ R, and for this we write x ⊥B y (cf. [6]). We refer to [10], [11], and [2] for basic properties of isosceles orthogonality and Birkhoff orthogonality, and to [4] and [3] for a detailed study of the relations between them.

For non-collinear rays [p, ai and [p, bi, the ray

· À 1³ a − p b − p ´ p, + + p 2 ka − pk kb − pk is called the Busemann angular¡ bisector of¢ the angle spanned by [p, ai and [p, bi, and it is denoted by AB [p, ai, [p, bi (cf. [8]). It is trivial to see that when ka − pk = kb − pk, then

¡ ¢ h 1 E A [p, ai, [p, bi = p, (a + b) . B 2 34 h. martini, s. wu

2. Some lemmas

The following lemmas are needed for our investigations.

Lemma 2.1. (cf. [1]) Let (M2, C) be a strictly convex Minkowski plane. Then, for any x ∈ (M2, C)\{O} and any number λ > 0, there exists a point y ∈ λC (unique except for the sign) such that x ⊥I y.

Lemma 2.2. Let C(x1, λ) and C(x2, λ) be two circles in a strictly convex 0 0 Minkowski plane (M2, C). If {w, z} ⊆ C(x1, λ), {w , z } ⊆ C(x2, λ), and w−z = w0 − z0, then 0 0 d(x1, hw, zi) = d(x2, hw , z i).

Proof. Without loss of generality, we can suppose that x1 = x2 = O and λ = 1. By the assumed strict convexity, kw − zk = kw0 − z0k = 2 implies that 0 0 0 0 w = −z and w = −z , yielding d(x1, hw, zi) = d(x2, hw , z i) = 0. Now we consider the case when kw − zk < 2. Again by strict convexity of (M2, C), any vector with norm < 2 is the sum of two unit vectors in a unique way (cf. [18, Proposition 14, p. 106]). Thus, either w = w0 and z = z0 or w = −z0 and 0 z = −w hold, and each of these two cases clearly gives that d(x1, hw, zi) = 0 0 d(x2, hw , z i).

Lemma 2.3. ([4, (4.12)]) If for any x, y ∈ (M2, C) with x ⊥I y there exists a number 0 < t < 1 such that x ⊥I ty, then (M2, C) is Euclidean. The following lemma follows immediately from Lemma 2.3.

Lemma 2.4. If for any x, y ∈ (M2, C) with x ⊥I y there exists a number t > 1 such that x ⊥I ty, then (M2, C) is Euclidean.

Lemma 2.5. (cf. [4, (10.2)]) A Minkowski plane (M2, C) is Euclidean if and only if the implication

x ⊥B y ⇒ x ⊥I y holds for any x, y ∈ C.

Lemma 2.6. (cf. [4, (10.9)]) A Minkowski plane (M2, C) is Euclidean if and only if the implication

x ⊥I y ⇒ x ⊥B y holds for any x, y ∈ C. on orthocentric systems 35

Lemma 2.7. (cf. [14]) Let {p1, p2, p3, p4} be a C-orthocentric system in a strictly convex Minkowski plane (M2, C). If xi is the circumcenter of the triangle pjpkpl, where {i, j, k, l} = {1, 2, 3, 4}, then {x1, x2, x3, x4} is also a C-orthocentric system and

pi − pj = xj − xi.

Lemma 2.8. Let (M2, C) be a strictly convex Minkowski plane. For any x, y ∈ (M2, C)\{O} with x ⊥I y, let p3 = y, p4 = −y, x1 = x, x2 = −x, and λ = kx + yk. Then there exist two points p1 ∈ C(x2, λ) and p2 ∈ C(x1, λ) such that {p1, p2, p3, p4} is a C-orthocentric system, and that one of the following conditions is satisﬁed:

(1) kp3 − p1k = kp3 − p2k, and p3 and the line hp1, p2i are separated by the line l0, which is passing through p4 parallel to hp1, p2i, £ ¤ p1+p2 (2) p4 ∈ p3, , 2 ¡ ¢ (3) p3 and the line hp1, p2i are separated by l0, and p4 ∈ AB [p3, p1i, [p3, p2i ,

(4) p3 and the line hp1, p2i are separated by the line l0, and hp1, p2i is a common supporting line of the circles C(x2, λ) and C(x1, λ).

Proof. (1) By the assumed strict convexity and the fact that x ⊥I y, one can easily verify that the circles C(x1, λ) and C(x2, λ) intersect in exactly two points, which are p3 and p4. Also, one can easily verify that 2x2 − y lies in the circle C(x2, λ) and 2x1 − y in C(x1, λ), and that the point p4 lies in the − segment [2x2 − y, 2x1 − y]. Denote by H the closed half plane bounded by − h2x2 − y, 2x1 − yi that does not contain p3, and by C(p4, λ) the intersection − of H and C(p4, λ). Then, since the point p3 and the line hx2 − 2y, x1 − 2yi are separated by the line h2x2 − y, 2x1 − yi, the points x1 − 2y and x2 − 2y lie − in the semicircle C(p4, λ) . Let 2 2 w = −y − x and z = −y + x. 3 3 Then simple calculation shows that [p3, x2 − 2y] intersects [2x2 − y, 2x1 − y] in w, and [p3, x1 − 2y] intersects [2x2 − y, 2x1 − y] in z. Since 2 1 2 kw − p k = kz − p k = kxk ≤ (kx + yk + kx − yk) = λ < λ, 4 4 3 3 3 for any t ∈ (0, 1) there exists a unique point x(t) such that the line hp3, tw + − (1 − t)zi intersects the semicircle C(p4, λ) in a point x(t). From the fact 36 h. martini, s. wu

x2 x1 O

2x2 − y wp4 z 2x1 − y

p1(t)(p1) p2(t)(p2) q x2 − 2y x1 − 2y

x(t)(x3)

Figure 1: The proof of Lemma 2.8.

that the segment [x1, x2 − 2y] intersects [x2, x1 − 2y] in p4 it follows that kx(t) − x1k < 2λ and kx(t) − x2k < 2λ hold for any t ∈ (0, 1) . Thus there exist two points p1(t) and p2(t) such that C(x(t), λ) intersects C(x2, λ) exactly in p1(t) and p4, and C(x(t), λ) intersects C(x1, λ) exactly in p2(t) and p4. Then, for any t ∈ (0, 1), {p1(t), p2(t), p3, p4} is a C-orthocentric system; see Figure 1. Moreover, for any t ∈ (0, 1) we have by Lemma 2.7 that p2(t) − p1(t) = x1 − x2. Then, by Lemma 2.2, ¡ ¢ ¡ ¢ d x(t), hp1(t), p2(t)i = d x2 − 2y, h2x2 − y, 2x1 − yi , and therefore ¡ ¢ ¡ ¢ d x(t), hp1(t), p2(t)i < d x(t), h2x2 − y, 2x1 − yi , which implies that p3 and the line hp1(t), p2(t)i are separated by the line h2x2 − y, 2x1 − yi. Now we show the existence of the points p1 and p2 with the desired properties. It is trivial that the functions x(t), p1(t), and p2(t) as well as the on orthocentric systems 37 function f(t) = kp3 − p2(t)k − kp3 − p1(t)k are continuous. So

lim f(t) = lim(kp3 − p2(t)k−kp3 − p1(t)k) = kp3 − (2x1 − y)k−kp3 − p4k > 0 t→0 t→0 and

lim f(t) = lim(kp3 − p2(t)k−kp3 − p1(t)k) = kp3 − p4k−kp3 − (2x2 − y)k < 0. t→1 t→1

Hence there exists a number t0 ∈ (0, 1) such that f(t0) = 0. Let x3 = x(t0), p1 = p1(t0), and p2 = p2(t0). Then p1 and p2 are two points having the desired properties. (2) For any t ∈ (0, 1), let the functions x(t), p1(t), and p2(t) be defined as in (1), and w(t), z(t) be the points where the line h2x2 − y, 2x1 − yi meets hp3, p1(t)i and hp3, p2(t)i, respectively. It is clear that when t is sufficiently close to 0, the midpoint of [w(t), z(t)] has to lie strictly between p4 and 2x1 −y, and when t is sufficiently close to 1, the midpoint of [w(t), z(t)] has to lie strictly between p4 and 2x2 − y. Thus there exists a number t0 ∈ (0, 1) such 1 ¡ ¢ that 2 w(t0) + z(t0) = p4. Then p1 = p1(t0) and p2 = p2(t0) are two points having the desired properties. (3) For any t ∈ (0, 1), let the functions x(t), p1(t), and¡ p2(t) be defined as in¢ (1). It is clear that when t moves¡ from 0 to 1, the¢ ray AB ¡[p3, p1(t)i, [p3, p2(t)i¢ turns continuously from AB [p3, 2x1 −yi, [p3, p4i to¡AB [p3, 2x2 −yi, [p3, p4¢i . Thus there exists a number t0 ∈ (0, 1) such that AB [p3, p1(t0)i, [p3, p2(t0)i = [p3, p4i. Let p1 = p1(t0) and p2 = p2(t0). Then p1 and p2 are two points having the desired property. 0 0 0 0 − (4) Let y ∈ C be a point such that y ⊥B x and {x2 +λy , x1 +λy } ⊆ H . 0 0 Then hx2 + λy , x1 + λy i is a common supporting line of the circles C(x2, λ) and C(x1, λ). 0 0 Let p1 = x2 + λy , p2 = x1 + λy , and x3 = p1 + p4 − x2. Then one can easily verify that {p1, p2, p4} ⊆ C(x3, λ), and therefore p1 and p2 are the two points with the desired properties.

3. Main Results

Now we will present our main results which are new characterizations of the Euclidean plane among all strictly convex normed planes via properties of C-orthocentric systems. 38 h. martini, s. wu

p3 = y

x2 x1 = x O

p4 p1 p2

Figure 2: The proof of Theorem 3.1.

Theorem 3.1. A strictly convex Minkowski plane is Euclidean if and only if for any C-orthocentric system {p1, p2, p3, p4} the relation

pi − pj ⊥B (pk − pl) holds, where {i, j, k, l} = {1, 2, 3, 4}.

Proof. If (M2, C) is Euclidean then, for any C-orthocentric system {p1, p2, p3, p4}, pi is the orthocenter of the triangle pjpkpl, where {i, j, k, l} = {1, 2, 3, 4}. Thus pi − pj ⊥B (pk − pl).

Conversely, for any x, y ∈ C with x ⊥I y let

p3 = y, p4 = −y, x1 = x, and x2 = −x.

By Lemma 2.8, there exist two points p1 and p2 such that {p1, p2, p3, p4} is a C-orthocentric system; see Figure 2. By Lemma 2.7, p2 − p1 = x1 − x2 = 2x. On the other hand, by the assumption of the theorem we have

(p2 − p1) ⊥B (p3 − p4) or, equivalently, x ⊥B y. By Lemma 2.6, (M2, C) is Euclidean. on orthocentric systems 39

Theorem 3.2. A strictly convex Minkowski plane (M2, C) is Euclidean if and only if for any C-orthocentric system® {p1, p2, p3, p4} with kp3 − p1k = p1+p2 kp3 − p2k it holds that p4 ∈ p3, 2 .

Proof. We only have to prove suﬃciency. By Lemma 2.4, we just need to show that for any x, y ∈ (M2, C) with x ⊥I y there exists a number t > 1 such that x ⊥I ty, and it is trivial in the case where at least one of x and y is O. For any x, y ∈ (M2, C)\{O} with x ⊥I y, let

p3 = y, p4 = −y, x1 = x, and x2 = −x.

By (1) of Lemma 2.8, there exist two points p1 and p2 such that {p1, p2, p3, p4} is a C-orthocentric system, kp3 − p1k = kp3 − p2k, and that p3 and the line hp1, p2i are separated by the line passing through p4 parallel to hp1, p2i. Then, by the assumption of the theorem, D p + p E p ∈ p , 1 2 . 4 3 2

Since p3 and the line hp1, p2i are separated by the line passing through p4 parallel to hp1, p2i, we have h p + p i p ∈ p , 1 2 . 4 3 2 Thus there exists a number t > 2 such that p + p t p − 1 2 = (p − p ) = ty. 3 2 2 3 4 On the other hand, by Lemma 2.7 we have ° ° °x − x ³ p + p ´° kx + tyk = ° 1 2 + p − 1 2 ° = kp − p k ° 2 3 2 ° 3 1 ° ° (3.1) °p − p ³ p + p ´° = ° 2 1 − p − 1 2 ° = kx − tyk . ° 2 3 2 °

Hence there exists a number t > 2 such that x ⊥I ty, which completes the proof.

Theorem 3.3. A strictly convex Minkowski plane (M2, C) is Euclidean if and only if for any C-orthocentric system {p1, p®2, p3, p4} the equality p1+p2 kp3 − p1k = kp3 − p2k holds whenever p4 ∈ p3, 2 . 40 h. martini, s. wu

The proof of Theorem 3.3 makes use of (2) of Lemma 2.8 and is very similar to that of Theorem 3.2, and so we omit it.

Theorem 3.4. A strictly convex Minkowski plane (M2, C) is Euclidean if and only if for¡ any C-orthocentric¢ system {p1, p2, p3, p4}, p4 lies on the line containing AB [p3, p1i, [p3, p2i whenever kp3 − p1k = kp3 − p2k.

Proof. We only have to prove sufficiency. By Theorem 3.2 it is sufficient® p1+p2 to show that for any C-orthocentric system {p1, p2, p3, p4}, p4 ∈ p3, 2 whenever kp3 − p1k = kp3 − p2k. By the assumption of the theorem, for any C-orthocentric system {p¡1, p2, p3, p4} with¢ kp3 − p1k = kp3 − p2k, p4 lies on the line containing AB [p3, p1i, [p3, p2i . By the definition of Busemann angular bisectors and the fact that kp3 − p1k = kp3 − p2k, we have ¡ ¢ h p + p E A [p , p i, [p , p i = p , 1 2 . B 3 1 3 2 3 2 ® ¡ ¢ p1+p2 Thus p3, 2 is the line containing AB [p3, p1i, [p3, p2i , and therefore D p + p E p ∈ p , 1 2 . 4 3 2 The proof is complete.

Theorem 3.5. A strictly convex Minkowski plane (M2, C) is Euclidean if and only if for any C-orthocentric system {p1, p2, p3, p4} the¡ equality kp3 − p¢1k=kp3 − p2k holds whenever p4 lies on the line containing AB [p3, p1i, [p3, p2i .

Proof. We only have to prove suﬃciency. By Lemma 2.4, we just need to show that for any x, y ∈ (M2, C) with x ⊥I y there exists a number t > 1 such that x ⊥I ty. For any x, y ∈ (M2, C)\{O} with x ⊥I y, let

p3 = y, p4 = −y, x1 = x, and x2 = −x.

By (3) of Lemma 2.8, there exist two points p1 and p2 such that {p1, p2, p3, p4} is a C-orthocentric system, p3 and the line hp1, p2i are separated by the line passing through p4 parallel to hp1, p2i, and that ¡ ¢ p4 ∈ AB [p3, p1i, [p3, p2i . on orthocentric systems 41

By the assumption of the theorem we have kp3 − p1k = kp3 − p2k, and then ¡ ¢ h p + p E A [p , p i, [p , p i = p , 1 2 . B 3 1 3 2 3 2

Since p3 and the line hp1, p2i are separated by the line passing through p4 parallel to hp1, p2i, we have h p + p i p ∈ p , 1 2 . 4 3 2 Hence there exists a number t > 2 such that p + p t p − 1 2 = (p − p ) = ty. 3 2 2 3 4 In a way analogous to that referring to Theorem 3.2 it can be proved that x ⊥I ty, which completes the proof.

Theorem 3.6. A strictly convex Minkowski plane is Euclidean if and only if for any C-orthocentric system {p1, p2, p3, p4} with p3 6= p4 the equality kp3 − p1k = kp3 − p2k holds whenever hp1, p2i is a common supporting line of the circle containing {p1, p3, p4} and the circle containing {p2, p3, p4}.

Proof. Suppose first that (M2, C) is Euclidean. For any C-orthocentric system {p1, p2, p3, p4}, let C(xi, λ) be the circumcircle of the triangle pjpkpl, where {i, j, k, l} = {1, 2, 3, 4}. Then hp3, p4i is the radical axis of C(x1, λ) and C(x2, λ). (Note that the radical axis of two circles is the locus of points of equal circle power with respect to two non-concentric circles, where the power of a point with respect to a circle is equal to the squared distance from the point to the center of the circle minus the squared radius of the circle; cf. [12].) If hp1, p2i is a common supporting line of C(x1, λ) and C(x2, λ), then hp3, p4i will be the perpendicular bisector of [p1, p2], and therefore kp3 − p1k = kp3 − p2k. Now we prove sufficiency. By Lemma 2.5 we only have to show that for any x, z ∈ C, z ⊥B x ⇒ z ⊥I x. Let ω be a fixed orientation on (M2, C). Since z ⊥B x if and only if (−z) ⊥B x, it will be sufficient to prove that for any −→ −→ x, z ∈ C with zx = ω (i.e., the orientation zx is given by ω), z ⊥B x implies z ⊥I x. By strict convexity of (M2, C), for any x ∈ C there exists a unique point −→ z ∈ C such that z ⊥B x and zx = ω. On the other hand, by Lemma 2.1, for any number t > 0 there is a unique point y(t) ∈ tC such that x ⊥ y(t) and −−−→ I y(t)x = ω. Let

x1 = x, x2 = −x, p3(t) = y(t), and p4(t) = −y(t). 42 h. martini, s. wu

p3 = y(t)

x2 x1 = x O

p4 p1(t) p2(t)

x3(t)

Figure 3: The proof of Theorem 3.6.

Then, by© (4) of Lemma 2.8, thereª exist two points p1(t) and p2(t) such that the set p1(t), p2(t), p3(t), p4(t) is a C-orthocentric system, p3(t) and the line hp1(t), p2(t)i are separated by the line passing through p4(t) which is parallel to hp1(t), p2(t)i, and that the line hp1(t), p2(t)i is the common supporting line of C(x1, kx + y(t)k) and C(x2, kx + y(t)k); see Figure 3. From Lemma 2.7 it follows that p1(t) − p2(t) = x2 − x1.

Thus x2 − p1(t) ⊥B x and x1 − p2(t) ⊥B x, and therefore

kx + y(t)k z = x2 − p1(t) = x1 − p2(t).

Let z(t) = kx + y(t)k z. Then

kp3(t) − p2(t)k = kp3(t) − x1 + x1 − p2(t)k = kp3(t) − x1 + z(t)k and

kp3(t) − p1(t)k = kp3(t) − x2 + x2 − p1(t)k = kp3(t) − x2 + z(t)k .

By assumption kp3(t) − p1(t)k = kp3(t) − p2(t)k, and therefore

kp3(t) − x2 + z(t)k = kp3(t) − p1(t)k = kp3(t) − p2(t)k = kp3(t) − x1 + z(t)k , on orthocentric systems 43 i.e., °¡ ¢ ° °¡ ¢ ° ° y(t) + z(t) + x° = ° y(t) + z(t) − x°. It is evident that lim y(t) = O, t→0 and therefore lim kx + y(t)k = kxk = 1. t→0 It follows that ¡ ¢ ¡ ¢ lim y(t) + z(t) = lim y(t) + kx + y(t)k z = z, t→0 t→0 and then ° ° ° ° ° ¡ ¢ ° ° ¡ ¢ ° kz + xk = °lim y(t) + z(t) + x° = °lim y(t) + z(t) − x° = kz − xk . t→0 t→0

Hence z ⊥I x.

The theorem above says that a Minkowski plane is Euclidean if and only if the implication

hp1, p2i supports the circles C(x1, λ) and C(x2, λ) =⇒ kp3 − p1k = kp3 − p2k holds for any C-orthocentric system {p1, p2, p3, p4}. In the next theorem we show that the reverse implication also characterizes the Euclidean plane.

Theorem 3.7. A strictly convex Minkowski plane is Euclidean if and only if for any C-orthocentric system {p1, p2, p3, p4} with p3 6= p4, hp1, p2i is a common supporting line of the circle containing {p1, p3, p4} and the circle containing {p2, p3, p4} whenever kp3 − p1k = kp3 − p2k.

Proof. For any x, z ∈ C with z ⊥B x and any number t > 0, we can deﬁne y(t), x1, x2, p3(t), and p4(t) as in the proof of Theorem 3.6. Then, by (1)© of Lemma 2.8, thereª exist two points p1(t) and p2(t) such that the set p1(t), p2(t), p3(t), p4(t) is a C-orthocentric system, p3(t) and the line hp1(t), p2(t)i are separated by the line passing through p4(t) parallel to hp1(t), p2(t)i, and kp3(t) − p1(t)k = kp3(t) − p2(t)k. By assumption, the line hp1(t), p2(t)i is the common supporting line of C(x1, kx + y(t)k) and C(x2, kx + y(t)k). Then, as in the proof of Theorem 3.6, it can be shown that z ⊥I x, which completes the proof. 44 h. martini, s. wu

Final remark. Since a real Banach space of dimension ≥ 2 is an inner product space if and only if each of its two-dimensional subspaces is isometric to the Euclidean plane, it is clear that all our theorems can be interpreted (in the spirit of the monograph [4]) as characterizations of inner product spaces among all strictly convex real Banach spaces of dimension ≥ 2.

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